Question

$$\begin{gathered} .05x+.2y=.07, \\ .3x-.1y=.03. \end{gathered}$$

Answer 1

The given equations are:
.05x + .2y = .07 ........(i)
.3x − .1y = .03 ..........(ii)

On multiplying (i) by 100 and (ii) by 100, we get:
5x + 20y = 7..........(iii)
30x − 10y = 3.........(iv)

On multiplying (iv) by 2, we get:
60x − 20y = 6 ..........(v)

On adding (v) from (vi), we get:
65x = 13
⇒ $x=\frac{13}{65}=\frac{1}{5}=0.2$

On substituting x = 0.2 in (iii), we get:
1 + 20y = 7
⇒ 20y = (7 − 1) = 6
⇒ $y=\frac{6}{20}=\frac{3}{10}=0.3$

Hence, the solution is x = .2 and y = .3.